00:01
To appropriately answer this question, we need to use a form of the avant -haf equation with the natural log of k is equal to negative delta h standard over r multiplied by 1 over the kelvin temperature plus delta s standard divided by r.
00:15
So if we plot the natural log of k as a function of 1 over the kelvin temperature, our slope is going to be equal to negative delta h divided by r.
00:25
If that's the case, then delta h will just be the negative of the slope times r.
00:30
Our y intercept will be equal to delta s standard over r.
00:34
And so delta s standard is just the y intercept multiplied by r.
00:39
So if we take the kelvin temperature and the equilibrium constant, take the reciprocal of the kelvin temperature and the natural log of the equilibrium constant and plot them, this is what we get for the slope, and this is what we get for the y intercept.
00:56
So 1 ,346 multiplied by negative rt, gives us our negative 11 ,190...